Problem: Complete the equation. $\dfrac34+ \dfrac34~=~$
Let's figure out what $\dfrac{3}{4}+ \dfrac{3}{4}$ equals. $\dfrac{0}{4}$ $\dfrac{3}{4}$ $\dfrac{6}{4}$ $\llap{{+}}\!\frac{3}{4}$ $\llap{{+}}\!\frac{3}{4}$ $\dfrac{3}{4}+ \dfrac{3}{4} = \dfrac{6}{4}$ Now, let's figure out how many times we add $\dfrac{1}{4}$ to make $\dfrac{6}{4}$. $\dfrac{0}{4}$ $\dfrac{1}{4}$ $\dfrac{2}{4}$ $\dfrac{3}{4}$ $\dfrac{4}{4}$ $\dfrac{5}{4}$ $\dfrac{6}{4}$ $\llap{{+}}\!\frac{1}{4}$ $\llap{{+}}\!\frac{1}{4}$ $\llap{{+}}\!\frac{1}{4}$ $\llap{{+}}\!\frac{1}{4}$ $\llap{{+}}\!\frac{1}{4}$ $\llap{{+}}\!\frac{1}{4}$ $=\overbrace{{\dfrac1{4}} +{\dfrac1{4}} +{\dfrac1{4}} + {\dfrac1{4}} + {\dfrac1{4}} + {\dfrac1{4}}}^{{6}\text{ fourths}} $ $=\dfrac{{6}\times{1}}{{4}}$ $\dfrac{3}{4}+ \dfrac{3}{4} = 6 \times \dfrac14$